Question

For a given medium, the polarizing angle is ${60}^{\circ }$. The critical angle for this medium is
$\left({\sin }^{-1}\right(0.5773)={35}^{\circ }{16}^{\prime }\text{and}{\sin }^{-1}(0.886)={60}^{\circ })$

• A. ${60}^{\circ }$
• B. ${35}^{\circ }{16}^{\prime }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }{30}^{\prime }$

Answer 1

The correct option is B ${35}^{\circ }{16}^{\prime }$

According to Brewster's law,
$\mu =\tan i_p$

$\Rightarrow \mu =\tan {60}^{\circ }=\sqrt{3}$

Also, from the condition of critical angle

$\sin i_c=\frac{1}{\mu }=\frac{1}{\sqrt{3}}=0.5773$

$\Rightarrow i_c={\sin }^{-1}\left(0.5773\right)$

$\Rightarrow i_c={35}^{\circ }{16}^{\prime }$

$\therefore$ $\left(B\right)$ is the correct option.

Why this Question? Key Concept: Brewster's law $\to \mu =\tan i_p$ Critical angle $\to \sin i_c=\frac{1}{\mu }$